docs: def vdist(Lat1: float, Lon1: float, Lat2: float, Lon2: float, ell: Ellipsoid = None) -> Tuple[float, float, float]:
      """
      Using the reference ellipsoid, compute the distance between two points
      within a few millimeters of accuracy, compute forward azimuth,
      and compute backward azimuth, all using a vectorized version of
      Vincenty's algorithm:
  
      ```python
      s,a12,a21 = vdist(lat1, lon1, lat2, lon2, ell)
      ```
  
      Parameters
      ----------
  
      Lat1 : float or numpy.ndarray of float
          Geodetic latitude of first point (degrees)
      Lon1 : float or numpy.ndarray of float
          Geodetic longitude of first point (degrees)
      Lat2 : float or numpy.ndarray of float
          Geodetic latitude of second point (degrees)
      Lon2 : float or numpy.ndarray of float
          Geodetic longitude of second point (degrees)
      ell : Ellipsoid, optional
            reference ellipsoid
  
      Results
      -------
  
      s : float or numpy.ndarray of float
          distance (meters)
      a12 : float or numpy.ndarray of float
          azimuth (degrees) clockwise from first point to second point (forward)
      a21 : float or numpy.ndarray of float
          azimuth (degrees) clockwise from second point to first point (backward)
  
  
  
      Original algorithm source:
      T. Vincenty, "Direct and Inverse Solutions of Geodesics on the Ellipsoid
      with Application of Nested Equations", Survey Review, vol. 23, no. 176,
      April 1975, pp 88-93.
      Available at: http://www.ngs.noaa.gov/PUBS_LIB/inverse.pdf
  
      Notes:
  
       1. lat1,lon1,lat2,lon2 can be any (identical) size/shape. Outputs will have the same size and shape.
       2. Error correcting code, convergence failure traps, antipodal corrections, polar error corrections, WGS84 ellipsoid parameters, testing, and comments: Michael Kleder, 2004.
       3. Azimuth implementation (including quadrant abiguity resolution) and code vectorization, Michael Kleder, Sep 2005.
       4. Vectorization is convergence sensitive; that is, quantities which have already converged to within tolerance are not recomputed during subsequent iterations (while other quantities are still converging).
       5. Vincenty describes his distance algorithm as precise to within 0.01 millimeters, subject to the ellipsoidal model.
       6. For distance calculations, essentially antipodal points are treated as exactly antipodal, potentially reducing accuracy slightly.
       7. Distance failures for points exactly at the poles are eliminated by moving the points by 0.6 millimeters.
       8. The Vincenty distance algorithm was transcribed verbatim by Peter Cederholm, August 12, 2003. It was modified and translated to English by Michael Kleder. Mr. Cederholm's website is http://www.plan.aau.dk/~pce/
       9. Distances agree with the Mapping Toolbox, version 2.2 (R14SP3) with a max relative difference of about 5e-9, except when the two points are nearly antipodal, and except when one point is near the equator and the two longitudes are nearly 180 degrees apart. This function (vdist) is more accurate in such cases.
          For example, note this difference (as of this writing):
  
          ```python
          vdist(0.2,305,15,125)
          ```
  
          > 18322827.0131551
  
          ```python
          distance(0.2,305,15,125,[6378137 0.08181919])
          ```
  
          > 0
  
       10. Azimuths FROM the north pole (either forward starting at the north pole or backward when ending at the north pole) are set to 180 degrees by convention.
           Azimuths FROM the south pole are set to 0 degrees by convention.
       11. Azimuths agree with the Mapping Toolbox, version 2.2 (R14SP3) to within about a hundred-thousandth of a degree, except when traversing to or from a pole, where the convention for this function is described in (10), and except in the cases noted above in (9).
       12. No warranties; use at your own risk.
      """
      if ell is None:
          ell = Ellipsoid()
  # %% prepare inputs
      lat1 = np.atleast_1d(Lat1)
      lat2 = np.atleast_1d(Lat2)
      lon1 = np.atleast_1d(Lon1)
      lon2 = np.atleast_1d(Lon2)
  
      assert lat1.shape == lon1.shape and lat2.shape == lon2.shape
  
      if lat1.shape != lat2.shape:
          if lat1.size == 1:
              lat1 = np.broadcast_to(lat1, lat2.shape)
              lon1 = np.broadcast_to(lon1, lon2.shape)
  
          if lat2.size == 1:
              lat2 = np.broadcast_to(lat2, lat1.shape)
              lon2 = np.broadcast_to(lon2, lon1.shape)
  # %% Input check:
      if ((abs(lat1) > 90) | (abs(lat2) > 90)).any():
          raise ValueError('Input latitudes must be in [-90, 90] degrees.')
  # %% Supply WGS84 earth ellipsoid axis lengths in meters:
      a = ell.a
      b = ell.b
  # %% preserve true input latitudes:
      lat1tr = lat1.copy()
      lat2tr = lat2.copy()
  # %% convert inputs in degrees to np.radians:
      lat1 = np.radians(lat1)
      lon1 = np.radians(lon1)
      lat2 = np.radians(lat2)
      lon2 = np.radians(lon2)
  # %% correct for errors at exact poles by adjusting 0.6 millimeters:
      kidx = abs(pi / 2 - abs(lat1)) < 1e-10
      if kidx.any():
          lat1[kidx] = sign(lat1[kidx]) * (pi / 2 - (1e-10))
  
      kidx = abs(pi / 2 - abs(lat2)) < 1e-10
      if kidx.any():
          lat2[kidx] = sign(lat2[kidx]) * (pi / 2 - (1e-10))
  
      f = (a - b) / a
      U1 = arctan((1 - f) * tan(lat1))
      U2 = arctan((1 - f) * tan(lat2))
      lon1 = lon1 % (2 * pi)
      lon2 = lon2 % (2 * pi)
      L = abs(lon2 - lon1)
      kidx = L > pi
      if kidx.any():
          L[kidx] = 2 * pi - L[kidx]
  
      lamb = L.copy()  # NOTE: program will fail without copy!
      lambdaold = np.zeros(lat1.shape)
      itercount = 0
      notdone = np.ones(lat1.shape, dtype=bool)
      alpha = np.zeros(lat1.shape)
      sigma = np.zeros(lat1.shape)
      cos2sigmam = np.zeros(lat1.shape)
      C = np.zeros(lat1.shape)
      warninggiven = False
      sinsigma = np.empty(notdone.shape)
      cossigma = np.empty(notdone.shape)
      while notdone.any():  # force at least one execution
          itercount += 1
          if itercount > 50:
              if not warninggiven:
                  logging.warning('Essentially antipodal points--precision may be reduced slightly.')
  
              lamb[notdone] = pi
              break
  
          lambdaold[notdone] = lamb[notdone]
  
          sinsigma[notdone] = sqrt(
              (cos(U2[notdone]) * sin(lamb[notdone]))**2 +
              (cos(U1[notdone]) * sin(U2[notdone]) - sin(U1[notdone]) *
               cos(U2[notdone]) * cos(lamb[notdone]))**2)
  
          cossigma[notdone] = (sin(U1[notdone]) * sin(U2[notdone]) +
                               cos(U1[notdone]) * cos(U2[notdone]) *
                               cos(lamb[notdone]))
          # eliminate rare imaginary portions at limit of numerical precision:
          sinsigma[notdone] = sinsigma[notdone].real
          cossigma[notdone] = cossigma[notdone].real
  
          sigma[notdone] = arctan2(sinsigma[notdone], cossigma[notdone])
  
          alpha[notdone] = (arcsin(cos(U1[notdone]) * cos(U2[notdone]) *
                                   sin(lamb[notdone]) / sin(sigma[notdone])))
  
          cos2sigmam[notdone] = (cos(sigma[notdone]) - 2 * sin(U1[notdone]) *
                                 sin(U2[notdone]) / cos(alpha[notdone])**2)
  
          C[notdone] = (f / 16 * cos(alpha[notdone])**2 *
                        (4 + f * (4 - 3 * cos(alpha[notdone])**2)))
  
          lamb[notdone] = (L[notdone] + (1 - C[notdone]) * f * sin(alpha[notdone]) *
                           (sigma[notdone] + C[notdone] * sin(sigma[notdone]) *
                            (cos2sigmam[notdone] + C[notdone] * cos(sigma[notdone]) *
                             (-1 + 2. * cos2sigmam[notdone]**2))))
          # print(f'then, lambda(21752) = {lamb[21752],20})
          # correct for convergence failure for essentially antipodal points
          if (lamb[notdone] > pi).any():
              logging.warning('Essentially antipodal points encountered. Precision may be reduced slightly.')
              warninggiven = True
              lambdaold[lamb > pi] = pi
              lamb[lamb > pi] = pi
  
          notdone = abs(lamb - lambdaold) > 1e-12
  
      u2 = cos(alpha)**2 * (a**2 - b**2) / b**2
      A = 1 + u2 / 16384 * (4096 + u2 * (-768 + u2 * (320 - 175 * u2)))
      B = u2 / 1024 * (256 + u2 * (-128 + u2 * (74 - 47 * u2)))
      deltasigma = (B * sin(sigma) *
                    (cos2sigmam + B / 4 * (cos(sigma) * (-1 + 2 * cos2sigmam**2) -
                                           B / 6 * cos2sigmam * (-3 + 4 * sin(sigma)**2) * (-3 + 4 * cos2sigmam**2))))
  
      dist_m = (b * A * (sigma - deltasigma))
  
  # %% From point #1 to point #2
      # correct sign of lambda for azimuth calcs:
      lamb = abs(lamb)
      kidx = sign(sin(lon2 - lon1)) * sign(sin(lamb)) < 0
      lamb[kidx] = -lamb[kidx]
      numer = cos(U2) * sin(lamb)
      denom = cos(U1) * sin(U2) - sin(U1) * cos(U2) * cos(lamb)
      a12 = arctan2(numer, denom)
      kidx = a12 < 0
      a12[kidx] = a12[kidx] + 2 * pi
      # %% from poles
      a12[lat1tr <= -90] = 0
      a12[lat1tr >= 90] = pi
      az = np.degrees(a12)
  
  # %% From point #2 to point #1
      # correct sign of lambda for azimuth calcs:
      lamb = abs(lamb)
      kidx = sign(sin(lon1 - lon2)) * sign(sin(lamb)) < 0
      lamb[kidx] = -lamb[kidx]
      numer = cos(U1) * sin(lamb)
      denom = sin(U1) * cos(U2) - cos(U1) * sin(U2) * cos(lamb)
      a21 = arctan2(numer, denom)
      kidx = a21 < 0
      a21[kidx] = a21[kidx] + 2 * pi
      # %% backwards from poles:
      a21[lat2tr >= 90] = pi
      a21[lat2tr <= -90] = 0.
      backaz = np.degrees(a21)
  
      return dist_m.squeeze()[()], az.squeeze()[()], backaz.squeeze()[()]